Understanding Array Bisection Algorithm

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2021-06-05

Understanding Array Bisection Algorithm

Given a sorted array arr, Array Bisection Algorithm (a.k.a. Bisection Method, Binary Search Method) enables us to find an insertion point i for a new element val such that arr[i-1] < val <= arr[i] (or, arr[i] < val <= arr[i+1]).

Problem

Consider we want to insert a number 10 to a sorted array [2, 4, 8, 16, 32]. Here, an insertion point should be i=3 as arr[2] < 10 <= arr[3].

A naive method is sequentially walking through the elements until we hit the condition:

def search(arr, val):
    """
    >>> search([2, 4, 8, 16, 32], 10)
    3
    """
    if val < arr[0]:
        return 0

    for i in range(1, len(arr)):
        if arr[i-1] < val and val <= arr[i]:
            return i

    return len(arr)

The time complexity of this approach is $O(N)$, and larger arrays take more time to complete the operation.

Intuition

As an optimized way to solve the problem, binary search finds out an insertion point in $O(\log N)$ time complexity.

The basic idea of the method is to repeatedly split arr into two chunks, first-half arr[:mid] and last-half arr[mid+1:] of the elements, until a dividing point mid reaches the target value val.

A GIF image below from penjee.com illustrates how it works in comparison with the naive method:

Implementation

Although Python implements the algorithm as a standard library bisect, let's try to implement it from scratch.

Starting from lo=0 and hi=len(arr)-1, what we have to do is to keep narrowing down a focused range while maintaining arr[lo] < val <= arr[hi].

def bisect(arr, val):
    """Bisection algorithm

    Return an index of an ascending-ordered array `arr` where `val` can be inserted. A returned index `i` indicates a potential insertion point, and
    `arr[i:]` must come after `val` once inserted.

    >>> bisect([2, 4, 8, 16, 32], 1)
    0
    >>> bisect([2, 4, 8, 16, 32], 4)
    1
    >>> bisect([2, 4, 8, 16, 32], 3)
    1
    >>> bisect([2, 4, 8, 16, 32], 10)
    3
    >>> bisect([2, 4, 8, 16, 32], 64)
    5
    """
    if len(arr) == 0:
        return 0
    if val < arr[0]:
        return 0
    if arr[-1] < val:
        return len(arr)

    lo, hi = 0, len(arr) - 1

    while lo < hi:
        if val == arr[lo]:
            return lo
        elif val == arr[hi]:
            return hi

        mid = (lo + hi) // 2

        if val == arr[mid]:
            return mid
        elif val < arr[mid]:
            hi = mid
        else:
            lo = mid + 1

    return lo

In the case of looking for a position where 10 fits in [2, 4, 8, 16, 32], the method updates lo and hi as follows.

First, all elements from head to tail are considered:

  2   4   8  16  32
  ^       ^       ^
  L       M       H

Next, the method realizes the first-half of the array elements is smaller than 10, and hence they are ignored so that the following process can focus only on the second half:

  2   4   8  16  32
              ^   ^
              M   H
              L

Finally, arr[2] < 10 <= arr[3] is confirmed, and 3 is returned as a potential insertion point:

  2   4   8  16  32
              ^
              H
              M
              L

Application

The bisection method can widely be applicable for searching a certain data point from historical records.

In real-world applications, it's safe to say that historical records arrive in the order of timestamp, and hence a target array is typically pre-ordered when we search something from there.

To give an example, assume you have a Tweet database for each user:

class User(object):

    def __init__(self):
        self.tweets = []

    def tweet(self, datetime, text):
        self.tweets.append((datetime, text))

The database sequentially stores a new tweet as soon as it's posted:

user = User()
user.tweet(20100101, 'Hello, world.')
# ...
user.tweet(20201201, 'I am hungry.')
user.tweet(20201231, 'Sleepy...')
user.tweet(202101015, 'Happy New Year!')
# ....

A question here could be "What was the last tweet in 2020?"

If we use bisect, an answer to the query can be easily and efficiently found by searching an insertion point for (20210101, ''):

def last_before(timestamp, arr):
    """
    arr[i] := (timestamp, value)
    """
    pos = bisect(arr, (timestamp, ''))
    if pos == 0:
        return ''
    if pos == len(arr):
        return arr[-1][1]
    if arr[pos][0] == timestamp:
        return arr[pos][1]
    return arr[pos-1][1]


last_before(20210101, user.tweets)  # => "Sleepy..."

Even if a target list is not pre-sorted, growing an array while maintaining its order is not hard when we leverage heap (sorted dictionary/queue, to be more precise). It only takes $O(\log N)$ for insertion.

The method itself is simple, but the efficient searching technique could accelerate a lot of real-life applications we can think of.

Author: Takuya Kitazawa

Takuya Kitazawa is a freelance software developer based in British Columbia, Canada. As a technologist specializing in AI and data-driven solutions, he has worked globally at Big Tech and start-up companies for a decade. At the intersection of tech and society, he is passionate about promoting the ethical use of information technologies through his mentoring, business consultation, and public engagement activities. See CV for more information, or contact at [email protected].

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